ABC is a right angled triangle. B being the right angle. Mid-points of BC and AC are respectively B' and A'. Area of `DeltaA'B'C'` is

Options

A. `1/2 xx "area of " Delta ABC `
B. `2/3 xx "area of " Delta ABC`
C. `1/4 xx "area of " Delta ABC`
D. `1/8 xx "area of " Delta ABC`

Answer

To find the area of triangle A'B'C, where A' and B' are the midpoints of sides AC and BC respectively in the right-angled triangle ABC (with B as the right angle), we can follow these steps:

### Step-by-Step Solution:

1. **Draw the Right-Angled Triangle**:
- Start by sketching triangle ABC with B as the right angle. Label the vertices A, B, and C.

2. **Identify the Midpoints**:
- Mark the midpoints of sides AC and BC. Let B' be the midpoint of BC and A' be the midpoint of AC.

3. **Use Properties of Similar Triangles**:
- Since A' and B' are midpoints, triangles A'B'C and ABC are similar. This is because they share angle C and both have a right angle (angle B and angle B').

4. **Determine the Ratio of Corresponding Sides**:
- The ratio of the lengths of the corresponding sides of similar triangles is equal to the ratio of the lengths of the segments formed by the midpoints. Since B' is the midpoint of BC, the length of B'C is half of BC. Similarly, A' is the midpoint of AC, making A'C half of AC.
- Therefore, the ratio of the sides A'B' to AB, B'C to BC, and A'C to AC is 1:2.

5. **Calculate the Area Ratio**:
- The area of similar triangles is proportional to the square of the ratio of their corresponding sides. Thus, if the ratio of the sides is 1:2, the area ratio will be (1/2)² = 1/4.

6. **Express the Area of Triangle A'B'C**:
- Let the area of triangle ABC be denoted as Area(ABC). Then, the area of triangle A'B'C can be expressed as:
\[ \text{Area}(A'B'C) = \frac{1}{4} \times \text{Area}(ABC) \]

### Final Result:
The area of triangle A'B'C is \(\frac{1}{4}\) times the area of triangle ABC.

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